Ltn→∞⁡1nf1n+f2n+……+f(2)=

limn→∞1n∑r = 124f(rn) = ∫12f(x)dx

Ltn→∞⁡1nf1n+f2n+……+f(2)=

limn→∞1n∑r = 124f(rn) = ∫12f(x)dx

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${Lt}_{n \to \infty} \frac{1}{n} \{f (\frac{1}{n}) + f (\frac{2}{n}) + \dots \dots + f (2)\} =$

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$\int_{0}^{1} f (\frac{1}{x}) dx$

$\int_{0}^{1} f (x) dx$

$\int_{0}^{1} f (2 x) dx$

$\int_{0}^{2} f (x) dx$

answer is D.

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$\lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{24} f (\frac{r}{n}) = \int_{1}^{2} f (x) dx$

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Ltn→∞⁡1nf1n+f2n+……+f(2)=

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${Lt}_{n \to \infty} \frac{1}{n} \{f (\frac{1}{n}) + f (\frac{2}{n}) + \dots \dots + f (2)\} =$